This rate of effusion calculator applies Graham’s law — a gas’s effusion rate is inversely proportional to the square root of its molar mass — to compare two gases: rate₁/rate₂ = √(M₂/M₁).

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Light Gases Leak Faster — by a Square Root

Effusion is a gas escaping through a hole so small that molecules pass one at a time without colliding — and its rate is set purely by how fast the molecules move. Temperature equalizes kinetic energy (½mv²) across gases, so speed must fall as mass rises: v ∝ 1/√M. Thomas Graham measured the result in the 1840s: the defaults compare helium (M = 4) against oxygen (M = 32), and He effuses √8 = 2.83× faster — which is why helium balloons sag overnight through pores that hold air for a week, and why a helium leak detector finds cracks nothing else can: no common gas escapes faster. The time form of the law runs the ratio the other way (time₂ = time₁ × √(M₂/M₁)): if helium empties a chamber in 60 seconds, oxygen needs ~170 seconds through the same pinhole.

The distinctions that keep this law honest: effusion (through a pinhole, molecule-by-molecule) is not bulk flow (through a pipe, pressure-driven — Graham’s law does not apply), and it is the single-molecule limit of diffusion, gases intermixing — related by the same √M physics, which is why perfumes (heavy molecules) spread slower than ammonia and why isotope separation worked at all: ²³⁵UF₆ and ²³⁸UF₆ differ by three neutrons on ~350 g/mol, a per-pass separation factor of √1.0086 ≈ 1.0043, compounded over thousands of stages into weapons- and reactor-grade enrichment. The square root is unforgiving — it compresses even a 100-fold mass difference into a 10-fold rate difference — which is both the law’s teaching point and the reason gaseous diffusion plants were measured in acres and gigawatts before centrifuges (which exploit mass directly, not its root) replaced them.

Worked Examples (Verified)

  • Defaults: He vs O₂: √(32/4) = 2.83× faster.
  • Hydrogen vs nitrogen: √(28/2) = 3.74× — the fastest common gas against air’s main component.
  • Times: gas 1 (M 4) effuses in 45 s → gas 2 (M 32) takes 45 × 2.83 = 127.3 s.
  • Unknown gas identification: a gas effusing at half O₂’s rate has √(M/32) = 2 → M = 128 g/mol (near Xe, 131).
  • UF₆ isotopologues: √(352/349) = 1.0043 per stage — thousands of stages required.

Effusion Discipline

  • Rate ratio uses the masses flipped: rate₁/rate₂ = √(M₂/M₁) — gas 1’s rate is set by gas 2’s mass. Getting the flip backwards is the classic error; sanity-check with “lighter is faster.”
  • Times invert: t₂/t₁ = √(M₂/M₁). Rates and times carry the same square root but answer opposite questions.
  • Same temperature is assumed — heat one gas and its rate rises by √(T₂/T₁) on top of the mass effect.
  • Graham’s law is exact for effusion (pinhole, mean-free-path ≫ hole); for diffusion through stagnant gas it’s the standard approximation, excellent for estimates.
  • Solving for an unknown molar mass: M_unknown = M_known × (rate_known/rate_unknown)² — the squared form this calculator’s ratio inverts cleanly.

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Frequently Asked Questions

How do you calculate the rate of effusion?

With Graham’s law: rate₁/rate₂ = √(M₂/M₁) — divide the two molar masses (second gas over first), take the square root, and you have how many times faster gas 1 effuses than gas 2. Helium (4) vs oxygen (32): √(32/4) = √8 = 2.83×. Note the inversion inside the root: a gas’s speed is governed by the OTHER gas’s mass in the ratio. This calculator also converts a known effusion time for gas 1 into the corresponding time for gas 2 (times scale by the same factor).

Why does helium escape balloons so fast?

Two Graham’s-law facts multiply: helium atoms are the second-lightest gas particles there are (M = 4 vs air’s ~29, so He moves √(29/4) ≈ 2.7× faster than air molecules), and latex is a porous membrane at the molecular scale whose pores pass helium readily while holding bulk air by its slower, bigger molecules (N₂ at 28, O₂ at 32). Mylar (foil) balloons use a metalized film with far smaller effective pores — same law, tighter holes, weeks instead of days of float.

What is the difference between effusion and diffusion?

Effusion is escape through a hole comparable to or smaller than the molecules’ mean free path: molecules pass independently, collide with nothing on the way out, and the rate follows Graham’s law exactly. Diffusion is the net spreading of one gas through another — molecules collide constantly and random-walk outward; its rate also scales roughly with 1/√M (same molecular-speed physics) but depends on the medium and concentration gradient too. Everyday language uses “diffusion” for both; the exam distinction is pinhole escape vs intermixing cloud.

How is Graham’s law used in uranium enrichment?

Uranium is converted to UF₆ gas, in which the fissile ²³⁵U molecules (M = 349) are slightly lighter than the ²³⁸U molecules (M = 352). Pumped against a porous barrier, the lighter isotopologue effuses √(352/349) ≈ 1.0043× faster, so the gas that passes is enriched by 0.43% per stage. Cascading thousands of stages compounds that whisper into 3–5% reactor fuel or far higher. The Oak Ridge K-25 plant — for a time the largest building on Earth — existed because that square root is so close to 1; gas centrifuges, which separate on the mass difference itself rather than its root, are roughly 50× more energy-efficient and replaced diffusion.

If one gas effuses in 60 seconds, how long does a gas twice as heavy take?

60 × √2 ≈ 84.9 seconds. Times scale with the square root of the mass ratio — double the molar mass costs √2 ≈ 1.41× the time, not double. Quadrupling the mass doubles the time. The calculator’s time field runs exactly this: enter gas 1’s molar mass, gas 2’s, and gas 1’s measured time, and it returns gas 2’s time through the same aperture at the same temperature and pressure difference.

Does temperature affect effusion rate?

Yes — molecular speed follows √T (kelvin) as well as 1/√M, so the complete scaling is rate ∝ √(T/M). Doubling absolute temperature raises a gas’s effusion rate by √2 ≈ 1.41×. Graham’s-law comparisons hold temperature constant (this calculator assumes it); when comparing measurements taken at different temperatures, fold the √(T₂/T₁) factor for each gas into the ratio before blaming the masses.

Can Graham’s law identify an unknown gas?

Yes — that’s one of its classical uses. Effuse the unknown and a known gas through the same apparatus and time them: M_unknown = M_known × (t_unknown/t_known)². A gas taking twice as long as O₂ (32 g/mol) to effuse the same volume has M = 32 × 4 = 128 g/mol — xenon territory (131). Combined with the density method (M = dRT/P), relative-effusion timing gave the 19th century two independent routes to molecular weights that corroborated each other.

Is this rate of effusion calculator free?

Yes — free, instant, no sign-up. Enter two molar masses for the effusion rate ratio, and optionally a time for gas 1 to get gas 2’s time.